A limit with Product and Cosine

Question

Can anyone help me please to find the limit of this series:

$$\lim_{x \to 0}\frac{\displaystyle\prod_{k=1}^{n}\left(1-\sqrt[k]{\cos x}\right)}{x^{2n-2}}$$

Thanks for all :D

Answer 1

As written, the limit is zero.

$$1-(\cos{x})^{1/k} = \frac{x^2}{2 k} + O(x^4)$$

so that the product is some factor times $x^{2 n}$ in this limit.

However,

$$\lim_{x \to 0}\frac{\prod_{k=1}^{k=n}1-\sqrt[k]{\cos x}}{x^{2n}} = \prod_{k=1}^{n} \frac{1}{2 k} = \frac{1}{2^n n!}$$

EDIT

Justifying the first step:

$$\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$$

$$\begin{align}(\cos{x})^{1/k} &= \left ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots \right )^{1/k} \end{align}$$

Now use the fact that $(1-z)^{1/k} = 1-\frac{z}{k}+\ldots$ and get

$$\begin{align}(\cos{x})^{1/k} &= \left ( 1 - \frac{x^2}{2!k} +\ldots \right )\end{align}$$

Note that the first term that I left out starts with $x^4$. Then we write, in place of the ellipses, $O(x^4)$ to represent the error which goes to zero as $x \rightarrow 0$. The first step then follows.

Answer 2

Note that by elementary limits $$\lim_{x \to 0}\frac{1-\sqrt[k]{\cos x}}{x^2}=\lim_{x \to 0}\frac{1-\sqrt[k]{\cos x}}{\ln \sqrt[k]{\cos x}}\times \frac{1}{k}\lim_{x \to 0}\frac{\ln \cos x}{x^2}=-1\times \left(-\frac{1}{2k}\right)=\frac{1}{2k}$$ and then

$$\lim_{x \to 0}\frac{\prod_{k=1}^{k=n}1-\sqrt[k]{\cos x}}{x^{2n}}=\prod_{k=1}^{k=n}\frac{1}{2k}=\frac{1}{2^n n!}$$

Chris.